Question

find the value alpha upon beta + beta upon alfa for 3x²+2x+5​

Answer 1

Given :-

The Quadratic equation 3x² + 2x + 5 = 0

To find :-

$\dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha }$

SOLUTION:-

As we know that ,

In the Quadratic equation,

If ${\alpha,\beta}$ are the roots then it is given by

$\alpha + \beta (sum \: of \: roots) = \dfrac{ - b}{a}$

$\alpha \beta(product \: of \: roots) = \dfrac{c}{a}$

So,

3x² + 2x + 5

Comparing with General form of Quadratic equation ax² + bx + c = 0

• a= 3
• b = 2
• c = 5

So,

$\alpha + \beta = \dfrac{ - 2}{3}$

$\alpha \beta = \dfrac{5}{3}$

Now, we need to find the

$\dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha }$

Take L.C.M to the denominators

$\dfrac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta }$

We shall do squaring on both sides for this equation in order to get the value of ${\alpha^2+\beta^2}$

$\alpha + \beta = \dfrac{ - 2}{3}$

$( \alpha + \beta ) {}^{2} = (\dfrac{ - 2}{3} ) {}^{2}$

$\alpha {}^{2} + \beta {}^{2} + 2 \alpha \beta = \dfrac{4}{9}$

$\alpha {}^{2} + \beta {}^{2} + \dfrac{2 \times 5}{3} = \dfrac{4}{9}$

$\alpha {}^{2} + \beta {}^{2} + \dfrac{10}{3} = \dfrac{4}{9}$

$\alpha {}^{2} + \beta {}^{2} = \dfrac{4}{9} - \dfrac{10}{3}$

$\alpha {}^{2} + \beta {}^{2} = \dfrac{4 - 30}{9}$

$\alpha {}^{2} + \beta {}^{2} = \dfrac{ - 26}{9}$

$\dfrac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta }$

Substituting the values ,

$\dfrac{ \dfrac{ - 26}{9} }{ \dfrac{5}{3} }$

$\dfrac{ - 26}{15}$

$\dfrac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta } = \dfrac{ - 26}{15}$

Answer 2

Given :-

The Quadratic equation 3x² + 2x + 5 = 0

To find :-

$\dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha }$

SOLUTION:-

As we know that ,

In the Quadratic equation,

If ${\alpha,\beta}$ are the roots then it is given by

$\alpha + \beta (sum \: of \: roots) = \dfrac{ - b}{a}$

$\alpha \beta(product \: of \: roots) = \dfrac{c}{a}$

So,

3x² + 2x + 5

Comparing with General form of Quadratic equation ax² + bx + c = 0

a= 3

b = 2

c = 5

So,

$\alpha + \beta = \dfrac{ - 2}{3}$

$\alpha \beta = \dfrac{5}{3}$

Now, we need to find the

$\dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha }$

Take L.C.M to the denominators

$\dfrac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta }$

We shall do squaring on both sides for this equation in order to get the value of ${\alpha^2+\beta^2}$

$\alpha + \beta = \dfrac{ - 2}{3}$

$( \alpha + \beta ) {}^{2} = (\dfrac{ - 2}{3} ) {}^{2}$

$\alpha {}^{2} + \beta {}^{2} + 2 \alpha \beta = \dfrac{4}{9}$

$\alpha {}^{2} + \beta {}^{2} + \dfrac{2 \times 5}{3} = \dfrac{4}{9}$

$\alpha {}^{2} + \beta {}^{2} + \dfrac{10}{3} = \dfrac{4}{9}$

$\alpha {}^{2} + \beta {}^{2} = \dfrac{4}{9} - \dfrac{10}{3}$

$\alpha {}^{2} + \beta {}^{2} = \dfrac{4 - 30}{9}$

$\alpha {}^{2} + \beta {}^{2} = \dfrac{ - 26}{9}$

$\dfrac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta }$

Substituting the values ,

$\dfrac{ \dfrac{ - 26}{9} }{ \dfrac{5}{3} }$

$\dfrac{ - 26}{15}$

$\dfrac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta } = \dfrac{ - 26}{15}$